Let F : R → R be a thrice differentiable function. Suppose that F 1 = 0 , F 3 = - 4 and F ′ x < 0 all x ∈ 1 2 , 3 . Let f x = x F ( x ) for all x ∈ R . If ∫ 1 3 x 2 F ′ x d x = - 12 and ∫ 1 3 x 3 F ′ ′ x d x = 40 , then the correct expression(s) is(are)

Step-by-step solution

∫ 1 3 x 2 F ′ x d x = - 12 a n d ∫ 1 3 x 3 F ′ ′ x d x = 40 ∫ 1 3 x 3 F ′ ′ x = x 3 F ′ x ] 1 3 - ∫ 1 3 3 x 2 F ′ ( x ) 40 = 27 F ′ 3 - F ′ 1 + 36 4 = 9 f 3 + 4 - f ( 1 ) ⇒ 9 f ′ 3 - f ′ 1 + 32 = 0 Also ∫ 1 3 x 2 F ′ ( x ) d x = x 2 F x ] 1 3 - ∫ 1 3 2 x F x d x = - 12 9 F 3 - F 1 - 2 ∫ 1 3 x F x d x = - 12 - 36 - 0 - 2 ∫ 1 3 f x d x = - 12 - 36 + 12 = 2 ∫ 1 3 f ( x ) d x - 12 = ∫ 1 3 f ( x ) d x